Can $\frac{a}{b} + \frac{b}{a}$ ever be bounded from above, if $a, b \in \mathbb{R}$? Good morning everyone!
By the arithmetic-geometric mean inequality, we all know that a suitable lower bound for the quantity
$$\frac{a}{b} + \frac{b}{a}$$
is $2$.
Now my question is:  Will this quantity ever be bounded from above, if we allow $a$ and $b$ to be in $\mathbb{R}$?
If the answer is NO, under what conditions on $a$ and $b$ can we ensure the existence of an upper bound?
 A: We have
$$ \lim_{a\rightarrow 0}\; \left|\frac{a}{b} + \frac{b}{a}\right| = \infty$$
and 
$$ \lim_{a\rightarrow \infty}\; \left|\frac{a}{b} + \frac{b}{a}\right| = \infty$$
regardless of your choice of $b$ (and vice-versa), so you won't be able to give an upper bound unless there is some relationship between $a$ and $b$.
A: Try $a = 1$ and $b \to \infty$. I would guess that it is bounded, iff $a/b$ and $b/a$ are bounded. So you don't win anything...
A: As most of the other answerers have noted, there is no global upper bound on $\frac xy + \frac yx$ because as either of them approaches 0, the term tends toward infinity.
However, your second part of the question has many different answers. For example, if you constrain $1 < x < 10$ and $1 < y < 10$, and then the upper bound will be $10.1$. In fact, you can even set a constraint $\frac xy + \frac yx < 3$ and the upper bound of that set will be $3$.
A: $(\dfrac{\sqrt a}{\sqrt b} - \dfrac{ \sqrt b}{\sqrt a})^2 \ge 0$
$\dfrac{a}{b} + \dfrac{b}{a} \ge 2$
Now you know it works for $a$ and $b$ being reals, too.
When you have to get the upper bound of the expression. Let one of the values to be maximum and other to minimum.
In this case:
$$\lim_{a\to \infty, b \to 0}\dfrac{a}{b} + \dfrac{b}{a} = \infty$$. Therefore upperbound of the expression is tending to $\infty$
